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# Test: Single Correct MCQs: Complex Numbers | JEE Advanced

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## 31 Questions MCQ Test Mathematics For JEE | Test: Single Correct MCQs: Complex Numbers | JEE Advanced

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Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 1

### If the cube roots of unity are 1, ω, ω2, then the roots of the equation (x – 1)3 + 8 = 0 are (1979)

Detailed Solution for Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 1

(x – 1)3  +  8 = 0
⇒ (x – 1)3   = – 8 = (– 2)3
⇒ x – 1 = – 2

or   -2ω or -2ω2

⇒ x = -1, 1 - 2ω, 1 -2ω2

Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 2

### The smallest positive integer n for which (1980)   = 1 is

Detailed Solution for Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 2

Now in = 1 ⇒ the smallest positive integral value of n should be 4.

Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 3

### The complex numbers z = x+ iy which satisfy the equation   lie on....... (1981 - 2 Marks)

Detailed Solution for Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 3

ATQ | x + iy – 5i | = | x + iy + 5i |
⇒ | x + (y – 5) i | = | x + (y + 5) i |
⇒ x2 + (y – 5)2 = x2 + (y + 5)2
⇒ x2 + y2 – 10y + 25 = x2 + y2 + 10y + 25
⇒ 20y = 0  ⇒ y = 0
∴ ‘a’ is the correct alternative.

Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 4

If   then                (1982 - 2 Marks)

Detailed Solution for Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 4

⇒ Re(z) < 0 and Im(z) = 0

∴ (b) is the correct choice.

Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 5

The inequality |z – 4| < |z –2| represents the region given by                 (1982 - 2 Marks)

Detailed Solution for Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 5

| z – 4 | <  | z – 2 |
⇒ | (x – 4) + iy | < | (x – 2) + iy |
⇒ (x – 4)2 + y2 < (x – 2)2 + y2
⇒ – 8x + 16 < – 4x + 4
⇒  4x – 12 > 0 ⇒ x  > 3 ⇒ Re (z) > 3

Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 6

If z = x + iy and ω = (1 - iz) /( z - i) , then |ω|= 1 implies that, in the complex plane,              (1983 - 1 Mark)

Detailed Solution for Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 6

⇒ |1– iz | = | z – i | ⇒ | 1 – i (x + iy) | = | x + iy – i |
⇒ | (y + 1) – ix | = | x + i (y – 1) |
⇒ x2 + (y + 1)2 = x2 + (y – 1)2
⇒ 4y = 0    ⇒ y = 0  ⇒ z lies on real axis

Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 7

The points z1, z2, z3 z4 in the complex plane are  the vertices of a parallelogram taken in order if and only if (1983 - 1 Mark)

Detailed Solution for Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 7

If vertices of a parallelogram are z1, z2, z3, z4 then as diagonals bisect each other

=  z1+ z3 = z2+ z4

Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 8

If a, b, c and u, v, w are complex numbers representing the vertices of two triangles such that c = (1 – r) a + rb and w = (1 – r)u + rv, where r is a complex number, then the two trian gles (1985 - 2 Marks)

Detailed Solution for Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 8

Let ABC be the Δ with vertices a, b, c and PQR be the Δ with vertices u, v, w.
Then c = (1– r) a + rb.

⇒ c – a = r(b – a) ....(1)

⇒ w = (1– r) u + rv  ....(2)

From (1) and (2) and

⇒ ΔABC ~ΔPQR.

Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 9

If ω (≠1) is a cube root of unity and (1 + ω)7 = A + B ω then A and B are respectively (1995S)

Detailed Solution for Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 9

(1 + ω)7 = A +Bω
⇒ (- ω2)7 = A +Bω (∵1 + ω +ω2 = 0)
⇒ - ω14 = A +Bω ⇒ - ω2 = A +Bω (∵ ω3= 1)
⇒ 1 + ω = A +Bω ⇒ A = 1, B = 1

Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 10

Let z and ω be two non zero complex numbers such that |z| = |ω| and Arg z+  Arggω = π, then z equals (1995S)

Detailed Solution for Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 10

Q | z | = | ω | and argz = π - argω

Let

Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 11

Let z and ω be two complex numbers such that |z| ≤1, |ω| ≤ 1 and |z + iω| = | z – i | = 2 then z equals (1995S)

Detailed Solution for Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 11

Given that

⇒ z lies on perpendicular bisector of the line segment joining which is real ax is , being mirror images of  each other..

∴ Im(z) = 0.

If z = x then

∴  (c) is the correct option.

Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 12

For positive integers n1, n2 the value of the expression (1 + i)n1 + (1 + i3)n1 + (1+ i5)n2 + (1+i7)n2 , where i = – is a real number if and only if                (1996 - 1 Marks)

Detailed Solution for Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 12

(1 + i) n1 + (1+ i3 )n1 + (1+ i5 )n2 + (1+i7 )n2

= (1 + i) n1 + (1 – i)n1 + (1 + i)n2+ (1 –i )n2

Using

and

We get the given expression as

= real number irrespective the values of n1 and n2

∴  (d) is the most appropriate answer.

Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 13

If i = , then 4 + 5  + 3 is equal to (1999 - 2 Marks)

Detailed Solution for Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 13

= 1+2w + 3(1+ w + w2) =

Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 14

If arg(z) < 0, then arg (-z) - arg(z) = (2000S)

Detailed Solution for Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 14

arg (z) < 0 (given) ⇒ arg (z) = – θ

Now

z = r cos(-θ) + i sin (-θ) = r [cos (θ) -i sin (θ)]
Again - z = -r [cos (θ) -i sin (θ)] = r [cos (π-θ) +i sin (π-θ)]
∴ arg (- z) = π -θ;
Thus arg (- z) - arg(z) =π - θ - (-θ) =π - θ +θ =π

Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 15

If z1, z2 and z3 are complex numbers such that (2000S)

|z1| = |z2| = |z3| =  = 1,   then   |z1 +z2 +z3|  is

Detailed Solution for Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 15

| z1 | = | z2 | = | z3| = 1  (given)

Now,

Similarly

Now,

NOTE THIS STEP

Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 16

Let z1 an d z2 be nth roots of unity which subtend a right angle at the origin . Then n must be of the form (2001S)

Detailed Solution for Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 16

Let z = (1)1/ n = (cos 2kπ + i sin 2kπ)1/n

k, 0, 1, 2, .....,n -1.

Let

and

be the two values of z. s.t. they subtend ∠ of 90° at origin.

As k1 and k2 are integers and k1 ≠ k2.

∴ n = 4k, k ∈ I

Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 17

The complex numbers z1, z2 and z3 satisfying   are the vertices of a triangle which is

Detailed Solution for Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 17

⇒ arg (cos(- π /3) + i sin (-π /3))
⇒ angle between z1 – z3  and  z2– z3  is 60°.

and

NOTE THIS STEP
⇒ The Δ  with vertices z1, z2 and z3 is isosceles with vertical ∠60° . Hence rest of the two angles should also be 60° each.
⇒  Req. Δ is an equilateral Δ.

Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 18

For all complex numbers z1, z2 satisfying |z1|=12 and | z2-3-4i| = 5, the minimum value of |z1-z2| is                        (2002S)

Detailed Solution for Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 18

| z1 |=12  ⇒ z1 lies on a circle with centre (0, 0) and radius 12 units, and | z2 – 3 – 4i | = 5
⇒ z2 lies on a circle with centre (3, 4) and radius 5 units.

From fig. it is clear that | z1– z2 | i.e., distance between z1 and z2 will be min when they lie at A and B resp. i.e., O,C, B, A are collinear  as shown.
Then z1– z2 = AB = OA – OB = 12 – 2(5) = 2.
As above is the min, value, we must have | z1– z2|  ≥ 2.

Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 19

If  |z| = 1 and  ω =   ( where Z ≠ 1) , then Re(ω) is

Detailed Solution for Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 19

Given that | z | = 1 and

Now we know that

(for  | z | = 1)

⇒ Re (ω)=0

Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 20

If ω(≠1) be a cube root of unity and (1 + ω2)n = (1 + ω4)n, then the least positive value of n is (2004S)

Detailed Solution for Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 20

(1 +ω2)n = (1 +ω4)n
⇒ (-ω)n = (1+ ω)n = (-ω2)n ⇒ ωn = 1 ⇒ n =3

Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 21

The locus of z which lies in shaded region (excluding the boundaries) is best represented by (2005S)

Detailed Solution for Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 21

Here we observe that.
AB = AC = AD = 2
∴ BCD is an arc of a circle with centre at A and radius 2.
Shaded region is outer (exterior) part of this sector ABCDA.
∴ For any pt. z on arc BCD we should have | z – (– 1) | = 2
and for shaded region, | z + 1| > 2 ....(i)
For shaded region we also have -p /4 < arg ( z + 1) <p /4
or | arg (z  + 1) | <p /4 ...(ii)

Combining (i) and (ii), (a)  is the correct option

Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 22

a, b, c are integers, not all simultaneously equal and  ω is cube root of unity (ω ≠ 1), then minimum value of |a + bω + cω2| is (2005S)

Detailed Solution for Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 22

Given that a, b, c are integers not all equal, w is cube root of unity ≠ 1,  then

| a + bω +cω2|

∴ The min value is obtained when any two are zero and third is a minimum magnitude integer i.e. 1.
Thus b = c = 0, a = 1 gives us the minimum value 1

Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 23

Let ω =- +i   then the value of the det.

(2002 - 2 Marks)

Detailed Solution for Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 23

Operating R1 + R2 + R3, we get

Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 24

If   is purely real where w = α + iβ, β ≠ 0 and z ≠ 1,then the set of the values of z is                       (2006 - 3M, –1)

Detailed Solution for Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 24

is purely real

⇒  | z |2  = 1 (∵ ω = α + iβ and β ≠ 0)

⇒  | z | = 1  also given z ≠ 1

∴ The required set is {z : | z | =1, z ≠ 1}

= 3ω (ω- 1)

Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 25

A man walks a distance of 3 units from the origin towards the north-east (N 45° E) direction. From there, he walks a distance of 4 units towards the north-west (N 45° W) direction to reach a point P. Then the position of P in the Argand plane is (2007 -3 marks)

Detailed Solution for Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 25

Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 26

If |z| = 1 and z ≠ ± 1, then all the values of  lie on                                                   (2007 -3 marks)

Detailed Solution for Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 26

Given | z | = 1 and z ≠ ± 1

To find locus of

We have

purely imaginary number

∴ ω must lie on y – axis.

Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 27

A particle P starts from the point z0 = 1 + 2i, where i = .It moves horizontally away from origin by 5 units and then vertically away from origin by 3 units to reach a point z1.From z1 the particle moves units in the direction of the vector and then it moves through an angle in anticlockwise direction on a circle with centre at origin, to reach a point z2. The point z2 is given by (2008)

Detailed Solution for Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 27

The initial position of point is Z0 = 1+ 2i
∴ Z1 = (1 + 5) + (2 + 3) i = 6 + 5i
Now Z1 is moved through a distance of 2 units in the direction

∴ It becomes

Now OZ1'  is rotated through an angle  inanticlockwise direction, therefore Z2 = iZ1' = - 6+7i

Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 28

Let z = cosθ + i sinθ. Then the value of

at θ  = 2° is (2009)

Detailed Solution for Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 28

z = cosθ + i sinθ

= cos(2m - 1)θ + i sin(2m - 1)θ

= sinθ + sin 3θ + sin 5θ + ......+ upto15 terms

Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 29

Let z = x + iy be a complex number where x and y are integers.Then the area of the rectangle whose vertices are the roots of the equation :                        (2009)

Detailed Solution for Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 29

Given z = x + iy where x and y are integer

Also

⇒ ( x2 +y2) ( x2 -y2)= 175
⇒ ( x2 +y2) ( x2 -y2) = 25x7 ...(i)
or(x2 + y2) (x2 – y2) = 35 × 5 ...(ii)
∵ x and y are integers,
∴ x2 +y2= 25 and x2 - y= 7       [From eq (i)]
⇒ x2 = 16 and y 2 = 9 ⇒ x = ± 4 and y =±3
∴ Vertices of rectangle are (4, 3) , (4, – 3), (– 4, – 3) , (– 4, 3).
So, area of rectangle = 8 × 6 = 48 sq. units Now from eq. (ii)
or x2 + y2 = 35 and x2 – y2 = 5
⇒ x2 = 20, which is not possible for any integral value of x

Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 30

Let z be a complex number such that the imaginary part of z is non-zero and a = z2  + z + 1 is real. Then a cannot take the value
(2012)

Detailed Solution for Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 30

∵ Im (z) ≠ 0
⇒ z is non real and equation z2 + z + (1 -a )=0 will have non real roots, if D < 0

⇒ 1 – 4(1 – a) < 0 ⇒ 4a < 3 ⇒ a <

∴ a can not take the value

Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 31

Let complex numbers a and  lie on circles (x – x0)2 + (y – y0)2 = r2 and (x – x0)2 + (y – y0)2 = 4r2. respectively. If z0 = x0 + iy0 satisfies the equation
2 |z0|2 = r2 + 2, then |α| =                                  (JEE Adv. 2013)

Detailed Solution for Test: Single Correct MCQs: Complex Numbers | JEE Advanced - Question 31

As α lies on the circle (x – x0)2 + (y – y0)2 = r2

∴ |α – z0|2  =  r2

(i)

Also  lies on the circle (x – x0)2 + (y – y0)2 = 4r2

Subtracting eqn (i) from (ii) we get

or

Using we get

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